What if I don't want to cancel out the 2 and I now have 1/2 integral 2xsin2x will it still give same answer

Here's why:
When you have a constant factor inside an integral, you can move it outside the integral sign.
So, $\int \frac{1}{2} \cdot 2x \sin(2x) \, dx$ is equivalent to $\frac{1}{2} \int 2x \sin(2x) \, dx$.

Let's re-examine the term from Step 1 of :

$$\int v \, du = \int \left(\frac{1}{2}\sin(2x)\right) (2x \, dx)$$

This is a product of three terms: $\frac{1}{2}$, $\sin(2x)$, and $2x$.
You can rearrange and factor out constants:

$$\int \frac{1}{2} \cdot 2x \sin(2x) \, dx = \int x \sin(2x) \, dx$$

Or, if you prefer to keep the $\frac{1}{2}$ and 2 separate initially:

$$\int \frac{1}{2} \cdot 2x \sin(2x) \, dx = \frac{1}{2} \int 2x \sin(2x) \, dx$$

When you then integrate $2x \sin(2x)$ and multiply by $\frac{1}{2}$, you will arrive at the same result as integrating $x \sin(2x)$ directly. It's just a different way of grouping the constants, but the mathematical outcome remains identical.